first off, let's notice the parabola is a vertical one, therefore the squared variable is the x, and the parabola is opening upwards, meaning the coefficient of x² is positive.
let's notice the vertex, or U-turn, is at (-2, 2)
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \boxed{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{-2}{ h},\stackrel{2}{ k}) \\\\\\ y=+1[x-(-2)]^2+2\implies y=(x+2)^2+2](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Cboxed%7By%3Da%28x-%20h%29%5E2%2B%20k%7D%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B-2%7D%7B%20h%7D%2C%5Cstackrel%7B2%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5C%5C%20y%3D%2B1%5Bx-%28-2%29%5D%5E2%2B2%5Cimplies%20y%3D%28x%2B2%29%5E2%2B2%20)
The answer to the equation is 17
Answer:
Step-by-step explanation:
=7 (1+11+111+1111......n)
=7/9 (9+99+999+9999....n)
=7/9 ((10-1)+(10^2-1)+(10^3-1)+....n)
=7/9 ((10+10^2+10^3...n)-(1+1+1+1.....n))
=7/9 ((10 (10^n-1)/(10-1))-n)
3.14 inches <span>is the length of the inscribed arc
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Looking at the given points on the right side from (0,2) to (1,6) for 1 increase in X ( 1-0=1) the Y value increases by 3 ( 6/2 = 3)
This same increase happens for th other two points: 18/6 = 3
54 / 18 = 3
The rate of increase is 3.
